PSLE In the figure, DEJK is a parallelogram and FGHJ is a rhombus. KJH is a straight line. ∠EDK = 58°, ∠JEF = 27° and ∠GFH = 33°.
- Find ∠EJF.
- Find ∠EFG.
(a)
∠EJK
= ∠EDK
= 58° (Parallelogram)
∠GHF
= ∠GFH
= 33° (Isosceles triangle)
∠FGH
= 180° - 33° - 33°
= 103° (Angles sum of triangle)
∠FJH
= ∠FGH
= 103° (Rhombus)
∠EJF
= 180° - 58° - 103°
= 19° (Angles on a straight line)
(b)
∠EFJ
= 180° - 27° - 19°
= 134° (Angles sum of triangle)
∠HFJ
= ∠GFH
= 33° (Rhombus)
∠EFG
= 360° - 134° - 33° - 33°
= 160° (Angles at a point)
Answer(s): (a) 19°; (b) 160°